I want to remove brackets and what is inside only if they are at the end of the string. Let's take three exemples :
s1 = 'BkToCstmrStmt/Stmt/Ntry[2]/NtryDtls/TxDtls/RltdPties/Cdtr/PstlAdr/AdrLine[2][]'
s2 = "BkToCstmrStmt/Stmt/Ntry[2]/AmtDtls/InstdAmt/Amt['CHF']"
s3 = "BkToCstmrStmt/Stmt/Bal[1]/Amt['CHF']"
I want to get
s1 = 'BkToCstmrStmt/Stmt/Ntry[2]/NtryDtls/TxDtls/RltdPties/Cdtr/PstlAdr/AdrLine'
s2 = "BkToCstmrStmt/Stmt/Ntry[2]/AmtDtls/InstdAmt/Amt"
s3 = "BkToCstmrStmt/Stmt/Bal[1]/Amt"
Here is what I tried :
name_parts = re.findall(r'[^\W_]+|[\W_]+', s3)
print(name_parts)
lenght = len(name_parts) - 1
# we want to analize the last element of the list, if it contains '_-'
if lenght >= 0: # it is to prevent an error if we have '' so an dimension '-1'
# We do a loop while to test if the parts have '-_', if true we execute the loop
# until it is false
while re.match('[^A-Za-z/]', name_parts[lenght]) or re.match('[^A-Za-z/]', name_parts[lenght-1]) :
# if it is true it will remove them
name_parts[lenght] = '' # it will remove them
print(name_parts)
lenght -= 1 # if the condition was true, we continue with one inferior part
else:
pass
new_string = ''.join(map(str, name_parts)) # now that we have cleaned if it was necessary
# we concatenate them
But It does not work. Anyone has an idea to efficiently do that ?